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POmysqli_error: updating tables (Procedural style)
 

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  1. POmysqli_error: updating tables (Procedural style)
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    <p>I am using PHP to try and update information I have in a mysqli table. I have decided to try and use mysqli rather than mysql. Unfortunately I cant seem to find my answer anywhere because im also trying to complete it Procedural style, as I have no knowledge of OOP and all tutorials (that i have found) are in OOP.</p> <p>Below is the script I have created. I have added comments to say what I think each command is doing.</p> <pre><code>&lt;?php DEFINE('DB_USER', 'root'); DEFINE('DB_PASS', 'password'); DEFINE('DB_NAME', 'test'); DEFINE('DB_HOST', 'localhost'); //connect to db $dbc = @mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME) or die(mysqli_connect_error($dbc)); mysqli_set_charset($dbc, 'utf8'); //form not submitted if(!isset($_POST['submit'])){ $q = "SELECT * FROM people WHERE people_id = $_GET[id]";//compares id in database with id in address bar $r = mysqli_query($dbc, $q);//query the database $person = mysqli_fetch_array($r, MYSQLI_ASSOC);//returns results from the databse in the form of an array }else{//form submitted $q = "SELECT * FROM people WHERE people_id = $_POST[id]";//compares id in database with id in form $r2 = mysqli_query($dbc, $q);//query the database $person = mysqli_fetch_array($r2, MYSQLI_ASSOC);//returns results from the database in an array $fname = $_POST['fname']; $lname = $_POST['lname']; $age = $_POST['age']; $hobby = $_POST['hobby']; $id = $_POST['id']; //mysqli code to update the database $update = "UPDATE people SET people_fname = $fname, people_lname = $lname, people_age = $age, people_hobby = $hobby WHERE people_id = $id"; //the query that updates the database $r = @mysqli_query($dbc, $update) or die(mysqli_error($r)); //1 row changed then echo the home page link if(mysqli_affected_rows($dbc) == 1){ echo "&lt;a href=\"index.php\"&gt;home page&lt;/a&gt;"; } } ?&gt; </code></pre> <p>The update form</p> <pre><code>&lt;form action="update.php" method="post"&gt; &lt;p&gt;First name&lt;input type="text" name="fname" value="&lt;?php echo "$person[people_fname]" ?&gt;" /&gt;&lt;/p&gt; &lt;p&gt;Last name&lt;input type="text" name="lname" value="&lt;?php echo "$person[people_lname]" ?&gt;" /&gt;&lt;/p&gt; &lt;p&gt;Your age&lt;input type="text" name="age" value="&lt;?php echo "$person[people_age]" ?&gt;" /&gt;&lt;/p&gt; &lt;p&gt;Your hobby&lt;input type="text" name="hobby" value="&lt;?php echo "$person[people_hobby]" ?&gt;" /&gt;&lt;/p&gt; &lt;input type="hidden" name="id" value="&lt;?php echo $_GET['id'] ?&gt;" /&gt; &lt;input type="submit" name="submit" value="MODIFY" /&gt; &lt;/form&gt;` </code></pre> <p>When I submit the form I get the following error message</p> <pre><code>Warning: mysqli_error() expects parameter 1 to be mysqli, boolean given in C:\xampp\htdocs\sandbox\update.php on line 39 </code></pre> <p>I realize this is telling me the issue is with</p> <p><code>$r = @mysqli_query($dbc, $update) or die(mysqli_error($r));</code></p> <p>So I have tried to put the sqli code in as the second parameter (i realize this is the same as putting the variable in, but it was a last resort), but it didn't seem right and still didn't work. I have also looked a php.net but couldn't work out the answer from the example they have given</p> <p>Please advise, I thought this was meant to be simple?</p>
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